Let $E/F$ be a finite field extension. Let $\text{Emb}(E/F)$ denote the set of field homomorphisms $E \to \overline{F}$ that fix $F$. Here, $\overline{F}$ is the algebraic closure of $F$.

My understanding is that the following things are true, but I haven't seen any explicit statements in the literature:

  • $E/F$ is normal $\iff$ $\text{Aut}(E/F) = \text{Emb}(E/F)$
  • $E/F$ is separable $\iff$ $|\text{Emb}(E/F)| = [E:F]$
  • In general: $|\text{Emb}(E/F)| \leq [E:F]$.

Is all of the above correct?

Second Question: Is there any relation of the above facts (if they're true) to the following one? (Dummit+Foote: Sec 14.1)

  • Fact: Let $\varphi \colon F \to F'$ be an isomorphism. Let $E$ be the splitting field of $f \in F[x]$, and $E'$ the splitting field of $\varphi(f) \in F'[x]$. Then there are $\leq [E:F]$ extensions of $\varphi$ to an isomorphism $E \to E'$. If $f$ is separable, then equality holds.
  • $\begingroup$ Should the right hand side of the first biconditional be $|\mathrm{Aut}(E/F)|=|\mathrm{Emb}(E/F)|$? $\endgroup$ – Julian Rosen Aug 3 '13 at 19:22
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    $\begingroup$ Not to nitpick, but it really doesn't make sense to compare $\mathrm{Aut}(E/F)$ and $\mathrm{Emb}(E/F)$, since an $F$-algebra homomorphism $E\to E$ is not the same sort of thing as an $F$-algebra homomorphism $E\to \overline{F}$. Personally, in this situation I would specify that $\overline{F}$ is an algebraic closure of $F$ that contains $E$, and then phrase the statement (which is true, by the way) that $E/F$ is normal $\iff$ every $F$-algebra homomorphism $E\to \overline{F}$ has image $E$ (and hence becomes an element of $\mathrm{Aut}(E/F)$ after restricting the codomain). $\endgroup$ – Zev Chonoles Aug 3 '13 at 19:23
  • $\begingroup$ I can't regard $\text{Aut}(E/F)$ as a subset of $\text{Emb}(E/F)$? $\endgroup$ – Jesse Qual Prep Aug 3 '13 at 21:04
  • $\begingroup$ @Jesse: I'm saying that part of the identity of a function is its domain and codomain - thus a collection of functions $A\to B$ can't possibly be equal to a collection of functions $C\to D$ unless $A=C$ and $B=D$. So unless $\overline{F}=E$, we can't have $\mathrm{Aut}(E/F)=\mathrm{Emb}(E/F)$. $\endgroup$ – Zev Chonoles Aug 3 '13 at 22:35
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    $\begingroup$ @Jesse: Here's the answer I mentioned: math.stackexchange.com/a/22531/264 $\endgroup$ – Zev Chonoles Aug 5 '13 at 4:06

All of the cited statements are true. You can find them (with different notation) in chapter $\mathsf{V}$ of Lang's Algebra, revised 3rd edition, which is on algebraic extensions.

From p.237-238:

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(proof of Theorem 3.3)

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From p.239-240:

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(argument that this doesn't depend on $\sigma$)

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(other stuff)

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Now for your second question, let $L$ be an algebraic closure of $F'$ containing $E'$. Note that $F$ and $E$ naturally have $F'$-algebra structures, via $\varphi^{-1}$. Let $\Phi:F\to L$ denote the map $\varphi:F\to F'$ considered as a map into $L$.

Any $F'$-algebra homomorphism $f:E\to L$ extending $\Phi$ produces a field $F'\subseteq f(E)\subseteq L$ in which $\Phi(f)=\varphi(f)$ splits. Because $E'$ is the splitting field of $\varphi(f)$ in $L$, we must have $f(E)\supseteq E'$; but $$[E:F]=[f(E):f(F)]=[f(E):F']$$ is equal to $[E':F']$ so that we would have to have $f(E)=E'$. Thus, any $F'$-algebra homomorphism $f:E\to L$ extending $\Phi$ will have image $E'$.

Therefore, an $F'$-algebra homomorphism $f:E\to L$ is equivalent to an $F'$-algebra homomorphism $E\to E'$ (by restricting or extending the codomain as necessary). Note that any $F'$-algebra homomorphism $E\to E'$ is necessarily an isomorphism, by a degree argument again.

Thus, the set of isomorphisms $E\to E'$ extending $\varphi$ has the same cardinality as the set of $F'$-algebra homomorphisms $f:E\to L$ extending $\Phi$, which is (almost, but not quite, by definition) equal to the separable degree $[E:F]_s$, which is less than or equal to $[E:F]$, and we have equality (by definition) when $E/F$ is separable, which is the case precisely when $f$ is separable.


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